Compute where is an infinite cyclic semigroup.
We showed in this previous exercise that every infinite cyclic semigroup is isomorphic to . We now prove a lemma.
Lemma: Let be a semigroup homomorphism and suppose generates . Then is determined uniquely by its behavior on in the sense that if is a semigroup homomorphism and for all , then . Proof: Since generates , every element of has the form where for some positive natural number . Now , so that .
In particular, is generated (additively) by 1. So every endomorphism of is determined uniquely by its image at 1.
On the other hand, if we define on by , we have . So is an endomorphism of . Since , by Lemma 1, if is an endomorphism of such that , then .
Define by . Note that, for all (additive), we have . So for all , and thus is a semigroup homomorphism. Certainly is surjective. Now if , then in particular, , so that . Hence is injective, and so an isomorphism.